You wouldn't power it at the 3.3V pin. You'd use the same pin you're using now, which I believe is called Vin on the nodeMCU. That's the pin that feeds into the nodeMCU's 3.3V regulator.
The problem is that the regulator needs the voltage on Vin to be higher than 3.3V to regulate properly. How much higher depends on the regulator, and it can vary quite a bit. Some regulators only need 3.5V, while others might need 4.5V. So it depends on what regulator is used on your nodeMCU. It's really not likely that your regulator has the very low dropout voltage you would need. On boards like these, they often use the cheapest regulators they can find, so you would have to test it to find out. If you have a variable power supply, you could see what voltage at Vin still works. Or just see how long one battery will work. You would be looking to see when the voltage on the 3.3V pin starts to drop.
Another alternative would be using a very low dropout external regulator, and feeding its output directly to the 3.3V pin of the nodeMCU. But that would involve a number of extra parts.
If you can find the regulator, you could post the markings on it, and maybe we can figure out what it is.
Edit: Actually, you could run the 4.2V from the battery(s) through a regular diode, like a 1N4001, to the 3.3V pin. You wouldn't have a regulator at all, but the diode would drop about 0.65V, which would bring the 4.2V down to 3.55V, which is barely within the 3.6V limit. This is not ideal because as the battery discharges, the diode will continue to drop the 0.65V. A regulator would only drop what it needs to drop to maintain 3.3V.